Two capacitors of capacitance 2C and C, respectively, are connected in series with an inductor of inductance L. Initially the capacitors have charge such that $V_{B} - V_{A} = 4 V_{0}$ and $V_{C} - V_{D} = V_{0}$ . Initial current in the circuit is zero. Find:
a. the maximum current that will flow in the circuit.
b. the potential difference across each capacitor at that instant.

I_{0} = V_{0} \sqrt{\frac{3 C}{L}}

; b.

V_{1} = 3 V_{0}, V_{2} = 3 V_{0}

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I_{0} = V_{0} \sqrt{\frac{6 C}{L}}

; b.

V_{1} = 3 V_{0}, V_{2} = 3 V_{0}

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I_{0} = V_{0} \sqrt{\frac{3 C}{L}}

; b.

V_{1} = 1.5 V_{0}, V_{2} = 3 V_{0}

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I_{0} = V_{0} \sqrt{\frac{6 C}{L}}

; b.

V_{1} = 1.5 V_{0}, V_{2} = 3 V_{0}

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Solution

The correct option is D $I_{0} = V_{0} \sqrt{\frac{6 C}{L}}$ ; b. $V_{1} = 3 V_{0}, V_{2} = 3 V_{0}$
Equivalent capacitance= $\frac{C_{1} C_{2}}{C_{1} + C_{2}} = \frac{2 C}{3}$
Total charge = $(2 C) (4 V_{0}) + (C) (V_{0}) = 9 C V_{0}$
Potential difference across A and D= $3 V_{0}$
From conservation of energy
$\frac{1}{2} L I^{2} = \frac{1}{2} C_{e f f} (3 V_{0})^{2} = \frac{1}{2} (\frac{2 C}{3}) (9 V_{0}^{2}) = 3 C V_{0}^{2}$
$⟹ I = V_{0} \sqrt{\frac{6 C}{L}}$
For maximum current to be present, the overall potential across A and D is zero.
$V_{1} + V_{2} = 0$
$\frac{Q_{1}}{2 C} = \frac{Q_{2}}{C}$
But conservation of charge must hold.
$Q_{1} + Q_{2} = 9 C V_{o}$
Hence charge across $2 C i s, Q_{1} = 6 C V_{0}$
Charge across $C i s, Q_{2} = 3 C V_{0}$
Hence potential across each capacitor $= 3 V_{0}$

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